Solve for p. 2(p + 1) = 18 8 8.5 9.5 10

Which statement is true?

True [87]

It’s the second one it’s right

8 0

3 years ago

Read 2 more answers

A figure can be covered by 28 unit squares,without any gaps or overlaps.what is the area of the figure?

Tom [10]

Remember that the area of a square is just one of its sides squared. $A=s^2$
where
$A$ is the area of the square
$s$ is one of the sides of the square

Unit square is a square whose sides have length 1, so lets use our formula to find the volume of one unit square:
$A=s^2$
$A=1^2$
$A=1$
We now know that each one of the squares has area 1 unit squared. Since there are 28 unit square in our figure, we are going to multiply the area of a unit square by 28 to find the area of our figure:
Area of the figure= $28(1)=28$ units squared

We can conclude that the area of the figure is 28 units squared.

7 0

3 years ago

Find an equation of the line that satisfies the given conditions. through (−5, −7); perpendicular to the line passing through (−

murzikaleks [220]

Answer:

The equation would be y = 2x + 3

Step-by-step explanation:

In order to solve this, we first need to find the slope of the line between (-2, 5) and (2, 3). In order to do this, we use the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (3 - 5)/(2 - -2)

m = -2/4

m = -1/2

Now that we have the original line with a slope of -1/2, we can tell a perpendicular line would have a slope of 2. This is because perpendicular lines have opposite and reciprocal slopes. Now we can use that slope and the given point in point-slope form to get the answer. Be sure to solve for y.

y - y1 = m(x - x1)

y + 7 = 2(x + 5)

y + 7 = 2x + 10

y = 2x + 3

7 0

3 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago

Factor 4x^2-36x+81 please help

SOVA2 [1]

(2x-9)^2... determine terms that are alike

8 0

3 years ago